PS2-1 - 13.15. CHAPTER 13, PROBLEM 15 25 13.15 Chapter 13,...

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Unformatted text preview: 13.15. CHAPTER 13, PROBLEM 15 25 13.15 Chapter 13, Problem 15 Problem: A ski jumper has an initial velocity v = V i and follows the indicated trajectory, landing on the inclined slope at time τ a distance from the beginning of the slope as shown in the figure. The slope is inclined to the horizontal at an angle θ. You can ignore effects of friction on the ski jumper’s motion. (a) Determine τ and as functions of V , g and θ, where g is gravitational acceleration. (b) What is the maximum vertical distance of the ski jumper from the slope, hmax ? Express your answer as a function of V , g and θ. (c) If θ = 30o and the ski jumper lands at τ = 3 sec, what are V , z . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . .................................................................................. .............................................................................. ...................................................................................... .. ............................................................................................... .......................................................................................... ... ........................................................................................................ ........................................................................................................ .................................................. ........................................................................................................ 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........................................................................................................ ........................................................................................................ ........................................................................................................ ........................................................................................................ and hmax ? v=V i x g = −g k θ So ut on: Because we gnore fr c on he mo on of he sk umper s sub ec on y o he cons an grav aona acce era on Deno ng me by t and e ng t = 0 when he sk umper aunches from he op of he s ope he equa ons and n a cond ons govern ng he mo on are d2 x =0 dt2 x(0) = 0 and x(0) = V d2 z = −g z (0) = 0 and z (0) = 0 dt2 (a) In egra ng w ce and mpos ng he n a cond ons he so u on o he equa ons for x and z s x(t) = V t When he sk umper ands a and 1 z (t) = − gt2 2 me τ we know ha x(τ ) = cos θ and z (τ ) = − s n θ τ= cos θ V 2 Therefore we have V τ = cos θ and 1 1 − s n θ = − gτ 2 = − g 2 2 Rearrang ng erms he so u ons for = 2V 2 tan θ g cos θ cos θ V =⇒ and τ n erms of V g and θ are as fo ows and τ= 2V tan θ g 26 CHAPTER 13. PARTICLE KINEMATICS (b) The vertical distance between the ski jumper and the slope surface is h = |z | − x tan θ = 12 gt − V t tan θ 2 To find the maximum value of h, we differentiate with respect to time, viz., dh = V tan θ − gt = 0 dt =⇒ t= V tan θ g The second derivative is d2 h/dt2 = −g < 0 so that this is indeed the time at which h achieves its maximum value. Hence, 2 V tan θ 1 V tan θ V 2 tan2 θ −V hmax = g tan θ = 2 g g 2g (c) Nothing in the problem statement specifies the units that should be use, so that we can use either USCS or SI. So, for the given value θ = 30o , we have the following. USCS Units: 32.174 ft/sec2 (3 sec) gτ ft V= = 83.6 = o 2 tan θ 2 tan 30 sec = 2(83.6 ft/sec)2 tan 30o 2V 2 tan θ = 289.6 ft = g cos θ 32.174 ft/sec2 cos 30o V 2 tan2 θ (83.6 ft/sec)2 tan2 30o = 36.2 ft = 2g 2 32.174 ft/sec2 hmax = SI Units: 9.807 m/sec (3 sec) m gτ = 25.5 = V= o 2 tan θ 2 tan 30 sec = 2(25.5 m/sec)2 tan 30o 2V 2 tan θ = 88.4 m = 2 g cos θ 9.807 m/sec cos 30o V 2 tan2 θ (25.5 m/sec)2 tan2 30o = = 11.0 m 2 2g 2 9.807 m/sec 2 hmax = ...
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This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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